maihuna wrote:i want to see efficient way to solve:

A B
X 3
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CA5

Certainly B must be 5. Since 3B = 15, we'll carry a 1 when we multiply by the second digit, so 3A + 1 = 10C + A, or 2A + 1 = 10C. The left side of that equation is odd, the right side even, so has no solutions. Is there a typo somewhere?

maihuna wrote: A B
X 6
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BBB

You might notice that 111 = 3*37, so any three digit number with identical digits is a multiple of 37. That only leaves two possibilities for AB: 37 or 74. We can quickly check that only 74 will give us the correct units digit, so must be right.